Squeeze Theorem/Functions/Proof 2
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Theorem
Let $a$ be a point on an open real interval $I$.
Let $f$, $g$ and $h$ be real functions defined at all points of $I$ except for possibly at point $a$.
Suppose that:
- $\forall x \ne a \in I: \map g x \le \map f x \le \map h x$
- $\ds \lim_{x \mathop \to a} \map g x = \lim_{x \mathop \to a} \map h x = L$
Then:
- $\ds \lim_{x \mathop \to a} \ \map f x = L$
Proof
Let $f, g, h$ be real functions defined on an open interval $\openint a b$, except possibly at the point $c \in \openint a b$.
Let:
- $\ds \lim_{x \mathop \to c} \map g x = L$
- $\ds \lim_{x \mathop \to c} \map h x = L$
- $\map g x \le \map f x \le \map h x$ except perhaps at $x = c$.
Let $\sequence {x_n}$ be a sequence of points of $\openint a b$ such that:
- $\forall n \in \N_{>0}: x_n \ne c$
and:
- $\ds \lim_{n \mathop \to \infty} \ x_n = c$
By Limit of Function by Convergent Sequences:
- $\ds \lim_{n \mathop \to \infty} \map g {x_n} = L$
and:
- $\ds \lim_{n \mathop \to \infty} \map h {x_n} = L$
Since:
- $\map g {x_n} \le \map f {x_n} \le \map h {x_n}$
it follows from the Squeeze Theorem for Real Sequences that:
- $\ds \lim_{n \mathop \to \infty} \map f {x_n} = L$
The result follows from Limit of Function by Convergent Sequences.
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): Appendix: $\S 18.6$: Limits of functions